ncert solutions physics class 11th

Gravitational potential (V) is constant at all points in a spherical shell. Hence the gravitational gradient ( $\frac{dV}{dr})$ is zero everywhere inside the spherical shell. The gravitational potential gradient is equal to the negative of gravitational intensity. Hence intensity is also zero at all points inside the spherical shell. This indicates that gravitational forces acting at a point in a spherical shell are symmetric

If the upper half of a spherical shell is cut out then the net gravitational force acting on a particle located at the centre O will be in the downward direction

Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction. Thus the gravitational intensity at centre O of the given hemispherical shell has the direction as indicated by arrow c.

(a) Legs hold the entire mass of a body in standing position due to gravitational pull. In space, an astronaut feels weightlessness because of the absence of gravity. Therefore swollen feet of an astronaut do not affect him/her in space

(b) A swollen face is caused generally because of apparent weightlessness in space. Sense organs such as eyes, ears, nose and mouth constitute a person's face. These symptoms can affect astronaut in space

(c) Headaches are caused because of mental strain. It can affect the working of an astronaut in space

(d) Space has different orientations. Therefore, orientational problem can affect an astronaut in space

(a) Total mechanical energy of a satellite is the sum of its kinetic energy (always positive) and potential energy (can be either positive or negative). At infinity, the gravitational potential energy of the satellite is zero. As the Earth-Satellite system is a bound system, the total energy of the satellite is negative. Thus the total energy of an orbiting satellite at infinity is equal to the negative of its kinetic energy.

(b) An orbiting satellite acquires a certain amount of energy that enables it to revolve around the Earth. This energy is provided by its orbit. It requires relatively lesser energy to move out of the influence of the Earth's gravitational field than a stationary object on the Earth's surface that initially contains no energy.

The rotation period of the satellite Io, $T_{ju}$ = 1.769 days = 1.769 x 24 x 60 x 60s

Radius of the Orbit is given by the relation, $R_{ju}$ = 4.22 $*{10}^8$ m

Mass is given by the relation: $M_j$ = $\frac{4{\pi }^2{R}_{ju}^3}{G{T}_{ju}^2}$ ……(i)

Where $M_j$ = mass of Jupiter and G = Universal gravitational constant

Orbital period of the Earth, $T_e$ = 365.25 days = 365.25 x 24 x 60 x 60 s

Orbital radius of the Earth, $R_e$ = 1 AU = 1.496 x ${10}^{11}$ m

Mass of Sun is given as : $M_s$ = $\frac{4{\pi }^2{R}_e^3}{G{T}_e^2}$ …(ii)

$\frac{M_s}{M_j}$ =$\frac{4{\pi }^2{R}_e^3}{G{T}_e^2}$ $*$ $\frac{G{T}_{ju}^2}{4{\pi }^2{R}_{ju}^3}$ =$\frac{R_e^3}{T_e^2}$ $*$ $\frac{T_{ju}^2}{R_{ju}^3}$= ( ${\frac{1.496\mathrm{}\mathrm{x}\mathrm{}{10}^{11}}{4.22\mathrm{}*{10}^8\mathrm{}})}^3$ $*({\frac{1.769\mathrm{}\mathrm{x}\mathrm{}24\mathrm{}\mathrm{x}\mathrm{}60\mathrm{}\mathrm{x}\mathrm{}60}{365.25\mathrm{}\mathrm{x}\mathrm{}24\mathrm{}\mathrm{x}\mathrm{}60\mathrm{}\mathrm{x}\mathrm{}60\mathrm{}})}^2$ = 1045.04

Hence it can be inferred that the mass of Jupiter is about one – thousandth that of the Sun